What exactly is this function doing? For a set S define the natural isomorphism Char: P(S) → F(S, {0, 1}).
Let A be a subset of S. if x is an element of A, then it goes to 1, if it isn't, it goes to 0. 
I drew a diagram on a paper to try to get a bigger understanding and larger picture of what's happening, but I just don't get it. I guess I think I can see that it's an isomorphism because the powerset of S has the same number of elements as the set of functions that go from subset of S to the element 1 but even this thought isn't fully secure in my head. But to be honest, I feel like this isomorphism is being forced to show something or to be used in someway. What way is that? 
Edit: Actually, how many functions Can go from a Set to two elements? I think I kind of assumed that since the domain is the powerset, it maps any subset whatsoever to the codomain, which is the set of all functions that go from a subset of S to {0,1}, which will be 1. So they do have the same cardinality in my head, but I'm still not too sure 
 A: Let us call this characteristic function by $\mathbb{1}$. Now $\mathbb{1}$ is a function from the power set of $S$ to $[S\rightarrow \{0,1\}]$ --- the set of functions from $S$ to $\{0,1\}$.
Therefore we might write $\mathbb{1}:\mathcal{P}(S)\rightarrow [S\rightarrow\{0,1\}]$, and the way it works is that the set $A\in\mathcal{P}(s)$ is sent to the function $\mathbb{1}_A$. This could be written as $A\mapsto \mathbb{1}_A$ or even $\mathbb{1}(A)=\mathbb{1}_A$ if you like the whole '$f(x)$' thing.
Now $\mathbb{1}_A\in[S\rightarrow\{0,1\}]$ and as you said this is defined by
$$\mathbb{1}_A(x)=\begin{cases}1 & \text{ if }x\in A\\ 0 & \text{ if }x\not\in A\end{cases}$$
...imagine your set $S$... now imagine a subset $A\subset S$. $\mathbb{1}_A$ 'paints' elements of $S$ --- black ($1$) if there are in $A$ or white ($0$) if they are not in $A$.
A: Here is a simple example which might clarify the situation.
Let $S = \{a,b,c\}$. This is our starting point.
Next, we form $\mathcal{P}(S)$, the set of all subsets of $S$. Explicitly:
$\mathcal{P}(S) = \{\emptyset, \{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},S\}$
Now let's look at $\text{Char}(A)$, for some sets $A \in \mathcal{P}(S)$.
Some easy ones:
$\text{Char}(\emptyset)$ is the following function:
$f:S \to \{0,1\}$ given by $f(s) = 0$ for all $s \in S$. Since the empty set has no elements, $f(s)$ is never $1$, so it is always $0$.
The other "obvious one" is $\text{Char}(S)$ which is the function:
$g:S \to \{0,1\}$ given by $g(s) = 1$ for all $s \in S$ -it's all $1$'s, since every $s \in S$ is in $S$.
Ok, so now let's look at a "less obvious" subset, say $A = \{a,b\}$. Which function might $\text{Char}(A)$ be? Let's call it $h$ for now.
$a \in A$, so $h(a) = 1$.
$b \in A$, so $h(b) = 1$.
$c \not\in A$, so $h(c) = 0$. So this is $h$:
$a \mapsto 1$
$b \mapsto 1$
$c \mapsto 0$.
For every possible domain element $s \in S$, we have $2$ choices of image for our "possible function" $S \to \{0,1\}$. Having chosen this for $a$, we have two such choices for $b$, and having chosen this, we have two choices for where to map $c$, giving us:
$2\cdot2\cdot2 = 8$ possible functions we can have. Strangely enough, there are $8$ possible subsets of $S$, that is $8$ elements in $\mathcal{P}(S)$.
